Question

A sample of a compound contains `"0.672 g Co"`, `"0.569 g As"`, and `"0.486 g O"`. What is its empirical formula?

Answer 1

The empirical formula is `"Co"_3"As"_2"O"_8`.

Explanation:

`color(white)(XX)"Atomic"`
`color(white)(XX)"Weight"color(white)(X)"Weight (g)"color(white)(X)"Wt./At. wt."color(white)(l)"Ratio"color(white)(Xll)"×2"`
`"Co"color(white)(Xl)58.9 color(white)(XXX)0.672color(white)(X)color(white)(XXX)0.0114color(white)(XXl)1.50color(white)(XX)3.00`
`"As"color(white)(Xl)74.9color(white)(XXX)0.569color(white)(XXXX)0.00760color(white)(XX)1color(white)(XXXl)2`
`"O"color(white)(Xll)16.0color(white)(XXX)0.486color(white)(XXXX)0.0304color(white)(XXll)4.00color(white)(XX)8.00`

First we have to calculate the ratio of the weight of each element to its atomic weight.

`"Co: 0.672/58.9 = 0.0114"`
`"As: 0.569/74.9 = 0.00760"`
`"O: 0.486/16.0 =" color(white)(ll)0.0304`

After calculating the ratio, find lowest value for the element, i.e. `0.00760`.

Divide this number into all the values above.

`"Co = 0.0114/0.0076 = 1.50"`
`"As = 0.0076/0.0076 = 1"`
`"O = 0.0304/0.0076 ="color(white)(ll)4.00`

Since the ratio for `"Co"` is not an integer, all ratios are multiplied by 2, which gives the empirical formula `"Co"_3"As"_2"O"_8`.

Answer 2

The empirical formula is `"Co"_3"As"_2"O"_8"` .

Explanation:

`0.672"g Co"`
`0.569"g As"`
`0.486"g O"`

Molar mass of each element (atomic weight in grams/mole)

`"Co":` `58.9332"g/mol"`
`"As":` `74.9216"g/mol"`
`"O":` `15.999"g/mol"`

Determine the number of moles of each element using the given mass and the molar mass.

`0.672"g Co"xx(1"mol Co")/(58.9332"g Co")="0.01152 mol Co"`

`0.569"g As"xx(1"mol As")/(74.9216"g As")="0.007594 mol As"`

`0.486"g O"xx(1"mol O")/(15.999"g O")="0.03038 mol O"`

Moles Ratios and Empirical Formula

Determine the mole ratios by dividing the number of moles of each element by the least number of moles.

`(0.01152)/(0.007594)"mol Co"=1.517`
`(0.007594)/(0.007594)"mol As"=1.00`
`(0.03038)/(0.007594)"mol O"=4.00`

An empirical formula is the lowest whole number ratio of elements in the compound. Problem: the mole ratio of Co is 1.5, which is not a whole number. In this case you multiply each ratio times 2.

The empirical formula is `"Co"_3"As"_2"O"_8"`.